What is the average value of the function #f(t) = t (sqrt (1 + t^2) )# on the interval #[0,5]#?
1 Answer
May 13, 2017
Explanation:
The average value of the function
#1/(b-a)int_a^bf(t)dt#
So here, the average value of
#1/(5-0)int_0^5tsqrt(1+t^2)dt#
Use the substitution
#=1/(5xx2)int_0^5sqrt(1+t^2)(2tdt)=1/10int_1^26sqrtudu=1/10int_1^26u^(1/2)du#
Using
#=1/10[u^(3/2)/(3/2)]_1^26=1/10(2/3)[u^(3/2)]_1^26=1/15(26^(3/2)-1^(3/2))=1/15(26sqrt26-1)#